A. Mafia

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/348/problem/A

Description

One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other n - 1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the i-th person wants to play ai rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?

Input

The first line contains integer n (3 ≤ n ≤ 105). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the i-th number in the list is the number of rounds the i-th person wants to play.

Output

In a single line print a single integer — the minimum number of game rounds the friends need to let the i-th person play at least ai rounds.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Sample Input

3
3 2 2

Sample Output

4

HINT

题意

有n个人,在玩一个游戏,游戏表示每局都必须有个管理员参加

告诉你,每个人想当多少局玩家,然后让你求,最少多少局游戏,才能满足题意!

题解:

贪心就好了,类似厨师煮饼那道题一样

max(max_num,sum/(n-1));

代码:

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 200001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** ll a[maxn];
ll sum=;
ll mx=;
int main()
{
int n=read();
for(int i=;i<=n;i++)
a[i]=read(),sum+=a[i],mx=max(a[i],mx);
n=n-;
ll ans;
ans=sum/n;
if(sum%n!=)
ans++;
cout<<max(mx,ans)<<endl;
}

Codeforces Round #202 (Div. 1) A. Mafia 贪心的更多相关文章

  1. Codeforces Round #202 (Div. 1) A. Mafia 推公式 + 二分答案

    http://codeforces.com/problemset/problem/348/A A. Mafia time limit per test 2 seconds memory limit p ...

  2. Codeforces Round #202 (Div. 2)

    第一题水题但是wa了一发,排队记录下收到的25,50,100,看能不能找零,要注意100可以找25*3 复杂度O(n) 第二题贪心,先找出最小的花费,然后就能得出最长的位数,然后循环对每个位上的数看能 ...

  3. Codeforces Round #202 (Div. 2) B,C,D,E

    贪心 B. Color the Fence time limit per test 2 seconds memory limit per test 256 megabytes input standa ...

  4. Codeforces Round #382 (Div. 2)B. Urbanization 贪心

    B. Urbanization 题目链接 http://codeforces.com/contest/735/problem/B 题面 Local authorities have heard a l ...

  5. Codeforces Round #164 (Div. 2) E. Playlist 贪心+概率dp

    题目链接: http://codeforces.com/problemset/problem/268/E E. Playlist time limit per test 1 secondmemory ...

  6. Codeforces Round #180 (Div. 2) B. Sail 贪心

    B. Sail 题目连接: http://www.codeforces.com/contest/298/problem/B Description The polar bears are going ...

  7. Codeforces Round #192 (Div. 1) A. Purification 贪心

    A. Purification Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/329/probl ...

  8. Codeforces Round #274 (Div. 1) A. Exams 贪心

    A. Exams Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/problem/A Des ...

  9. Codeforces Round #374 (Div. 2) B. Passwords 贪心

    B. Passwords 题目连接: http://codeforces.com/contest/721/problem/B Description Vanya is managed to enter ...

随机推荐

  1. oracle数据库只查询前n条

    select * from  (select * from   tablename order by createdate desc)  aaa -- 按创建时间倒排序 where rownum &l ...

  2. numpy 简介

    .caret, .dropup > .btn > .caret { border-top-color: #000 !important; } .label { border: 1px so ...

  3. dpkg的用法 (转)

    dpkg是一个Debian的一个命令行工具,它可以用来安装.删除.构建和管理Debian的软件包. 下面是它的一些命令解释: 1)安装软件 命令行:dpkg -i <.deb file name ...

  4. Redis使用详细教程【转】

    转自 Redis使用详细教程 - wangyuyu - 博客园http://www.cnblogs.com/wangyuyu/p/3786236.html 一.Redis基础部分: 1.redis介绍 ...

  5. HOJ 1108

    题目链接:HOJ-1108 题意为给定N和M,找出最小的K,使得K个N组成的数能被M整除.比如对于n=2,m=11,则k=2. 思路是抽屉原理,K个N组成的数modM的值最多只有M个. 具体看代码: ...

  6. Freemaker 自定义指令和函数

    自定义函数和指令都可以在前台或者后台进行指定. 个人理解:指令的作用,主要是进行页面调整之后进行输出:函数的作用,主要是为了进行运算,返回运算结果供前台展示. (一) 自定义指令 使用以下格式调用自定 ...

  7. ssh使两台机器建立连接

    ssh利用口令建立连接过程: 客户端--> 发送连接请求 --> 远程主机 --> 返回远程主机的公钥 --> 公钥加密客户端私钥+客户端公钥返回远程主机 --> 远程主 ...

  8. CentOS系统yum源配置修改、yum安装软件包源码包出错解决办法apt.sw.be couldn't connect to host

    yum安装包时报错: Could not retrieve mirrorlist http://mirrorlist.repoforge.org/el6/mirrors-rpmforge error  ...

  9. 窗口生效函数UpdateData

    Invalidate()使整个窗口客户区无效.窗口的客户区无效意味着需要重绘,例如,如果一个被其它窗口遮住的窗口变成了前台窗口,那么原来被遮住的部分就是无效的,需要重绘.这时Windows会在应用程序 ...

  10. BeanUtils封装对象时一直提示ClassNotFoundException:org.apache.commons.beanutils.BeanUtils

    导包明明正确了,依赖包也全都导对了,还是出错. 困扰了3天. 后来看到这篇博文,https://blog.csdn.net/yanshaoshuai/article/details/81624890 ...